Analysis of rolling

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Analysis of rolling

  1. 1. MAE 212: SLAB ANALYSIS FOR FLAT ROLLING N. Zabaras, Spring 2000 Figure 1: Schematic of flat rolling showing the neutral point, Æ . ¼ Ü µ § · ´ Ü· ܵ ´ · ´¾ µ ÔÊ ¾ÔÊ Ü · Ü· ßÞ Ü   Ü Ü Ó× (1) ÒØÖÝ · ·   ×Ò   Ü µ ÜØ §¾ ÔÊ Ó× · ¾ÔÊ Þ ÖÓ ´ Ü µ ¼  × Ò ¦ ¾ÔÊ ´ Small angles: µ Ó× µ ¿ µ ×Ò Ó× ´ Ü µ 1 ×Ò ½   ¦ ¾ÔÊ ´ µ (2)
  2. 2. Figure 2: Stresses on an element in rolling: (a) entry zone and (b) exit zone. For small angles take Þ ³  Ô and for plane strain ´¯Ý ¼µ µ ¾ Ô Ü·Ô ßÞ (Y = Yield Stress) ¿ (3) true anywhere inside the deformation zone changes with as follows: Figure 3: Approximation of in terms of . · ¾ ´Ê µ × Ò · ¾ ´Ê µ ·Ê 2 ¾ ¾ ¾ (4)
  3. 3. We assume that as the material advances inside the deformation zone, its hardening behavior is ÓÒ×Ø ÒØ (so as decreases, increases such that the product remains such that: constant!!! A ridiculous assumption that however is better than assuming that is constant inside the deformation zone!). Returning to the equilibrium equation with the above assumption, we can write:  Ô ¾ Ô¿ ´ Ü µ ¾ Ô ¿ Ô   ´ Ô½   µ ´ µ ßÞ   ´Ô µ Þ ÖÓ   ´Ô µ (5) ßÞ ÓÒ×Ø Finally: Ô   ÔÊ ´  ¦ µµ ¾ Ô   Ê ¾ Ô   ¦ µ Ô Ê  ·¦ Ê   ´Ô µ ¾ « to a general angle . Similar Let us integrate the above equation in the entry region from calculation can be applied to the exit region.     · ÐÒ ¾ Ô Ô ÒØÖÝ   ÐÒ Ô Ê Õ ¼ ½ ¾ Ô Ê Ø ÒØÖÝ × Ê   ÒØÖÝ  ÐÒ  ½ Ê Ò   · Ø µ · · Ê ¾ Ê ¾ × (6) ¾ (7) · ÐÒ  ½ Ê « Ò · ½ Ê«¾ (8) Note that in the last calculation we used the following integral formula: Ü ¾ · ¾ ܾ ½ Ö Ø Ò 3 Ü ´ Ö Ø Ò  ½ µ Ø Ò (9)
  4. 4. At the entry region using the yield condition, one can write the following: p |entry = Y where Yentry = 2Y √ 3 − σx Y 2 σb √ − 3 Yentry |entry = 2 σb = √ 1− Yentry 3 (10) 2 √ Yentry . 3 So returning to equation (8), we can write: −ln σb p 2 + ln √ 1 − Y Yentry 3   h   ho = −ln hf + Rφ2  + ln hf + Rα2       1 + 2Rµ  R R  φ − tan−1 α h h −1 tan hf R (11) Define: R tan−1 hf H = 2 Ho  R φ hf entry  R R = 2 tan−1  α  hf hf (12) Equation (11) is now simplified as: −ln −ln ln p Y 2 √ 3 1− p Y h 1− p Y h 1− ho = µ (H − Ho ) ⇒ σb Yentry ho h = µ (H − Ho ) σb Yentry ho = µ (Ho − H) σb (13) Yentry Finally, the following pressure distribution is derived in the entry region: p σb = 1− Y Yentry 4 h µ(Ho −H) e ho (14)
  5. 5. where H and Ho are given by equation (12). To derive the corresponding equation in the exit region, you can repeat the above calculations by integrating equation (6) (with the bottom sign in ±) from angle φ to angle 0 (exit). It is also possible to derive the distribution of p at the exit using equation (14) with some changes! here ho → hf , Yentry → Yexit (15) Ho → 0 (because α = φat the exit = 0) σf p = 1− Y Yexit h µH e hf (16) Equations (14) and (16) define the complete pressure distribution in the deformation zone. Calcuation of the Neutral Point Equate the two pressure expressions from equations (14) and (16): 1− σb Yentry σf h µ(Ho −H) e = 1− ho Yexit µ(2H−Ho ) ⇒e = 1− 1 σb Yentry σ −Yf exit h µH e hf hf ho (17) (18) Simplifying for the case σb = σf = 0 leads to: Hn = 2 1 1 ho Ho − ln 2 µ hf R tan−1 hf φn = hf R (19) R φn = Hn ⇒ hf (20) hf Hn R 2 (21) tan 5

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